Answer: The given information results in a two triangles.
The first one with sides ;
a = 11.4, b = 6 and c = 8
Angle A = 108, B = 30 and C = 42
The second one with sides;
a = 10, b = 6 and c = 8
Angle A = 90, B = 30 and C = 60.
Step-by-step explanation: With the information provided, we can find the size of the angle opposite the third side which we shall call side a, and then calculate the length of side a. We use the sine rule which states that
a/SinA = b/SinB = c/SinC
6/Sin 30 = 8/SinC
By cross multiplication we now have
SinC = (8 x Sin 30)/6
SinC = (8 x 0.5)/6
SinC = 4/6
SinC = 0.6667
C = 41.8128
Approximately, angle C is 42 degrees.
That makes angle A to be
A = 180 - (42 + 30) {Sum of angles in a triangle equals 180}
A = 108
Therefore to calculate length a using the sine rule;
a/SinA = b/SinB
a/Sin 108 = 6/Sin 30
a = (6 x Sin 108)/Sin 30
a = (6 x 0.9510)/0.5
a = 5.706/0.5
a = 11.412
Approximately line a equals 11.4 units
Therefore the sides of the first triangle becomes 6, 8 and 11.4 units while it’s angles are 30, 42 and 108 degrees.
Also, if two sides are given as 6 units and 8 units respectively, then the third side can be calculated by using the Pythagoras theorem and this immediately presumes that it’s a right angled triangle (one of the angles equals 90 degrees). The theorem states that;
AC^2 = AB^2 + BC^2
Where AC is the hypotenuse/longest side. Taking the other two as 6 and 8 units, the formula now becomes;
AC^2 = 6^2 + 8^2
AC^2 = 36 + 64
AC^2 = 100
Add the square root sign to both sides of the equation
AC = 10.
Therefore the sides of the second triangle becomes 6, 8 and 10 units with angles as 30, 60 and 90 degrees.
